A converging lens of focal length 12 cm and a diverging mirror of focal length 7.5 cm are placed 5.0 cm apart with their principal axes coinciding. Where should an object be placed so that its image falls on itself?

#### Solution

Let the object be placed at a distance *x *cm from the lens (away from the mirror).

For the convex lens (1^{st} refraction) *u* = − *x*, *f* = − 12 cm

From the lens formula:

\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\Rightarrow\frac{1}{v}=\frac{1}{( - 12)}+\frac{1}{( - x)}\Rightarrow v=-\left( \frac{12x}{x + 12} \right)\]

Thus, the virtual image due to the first refraction lies on the same side as that of object A'B'.

This image becomes the object for the convex mirror,

For the mirror,

\[u = - \left( 5 + \frac{12x}{x + 12} \right)\]

\[ = - \left( \frac{17x + 60}{x + 12} \right)\]

\[f = - 7 . 5 \text{ cm }\]

From mirror equation,

\[\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\]

\[ \Rightarrow \frac{1}{v} = \frac{1}{- 7 . 5} + \frac{x + 12}{17x + 60}\]

\[ \Rightarrow \frac{1}{v} = \frac{17x + 60 - 7 . 5}{7 . 5(17x + 60)}\]

\[\Rightarrow v = \frac{7 . 5(17x + 60)}{52 . 5 - 127 . 5x}\]

\[ \Rightarrow v = \frac{250(x + 4)}{15x - 100}\]

\[ \Rightarrow v = \frac{50(x + 4)}{(3x - 20)}\]

Thus, this image is formed towards the left of the mirror.

Again for second refraction in concave lens,

\[u = - \left[ \frac{5 - 50(x + 4)}{3x - 20} \right]\]

(assuming that the image of mirror formed between the lens and mirror is 3*x* − 20),*v* = + *x* (since, the final image is produced on the object A"B")

Using lens formula:

\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]

\[ \Rightarrow \frac{1}{x}+\frac{1}{\frac{\left[ 5 - 50 (x \times 4) \right]}{3x - 20}}=\frac{1}{- 20}\]

⇒ 25*x*^{2} − 1400*x* − 6000 = 0

⇒ *x*^{2} − 56*x* − 240 = 0

⇒ (*x* − 60) (*x* + 4) = 0

Thus, *x* = 60 m

The object should be placed at a distance 60 cm from the lens farther away from the mirror, so that the final image is formed on itself.